A car accelerates uniformly from rest and
reaches a speed of 29.1 m/s in 6.5 s. The
diameter of a tire is 83.4 cm.
Find the number of revolutions the tire
makes during this motion, assuming no slipping.
Answer in units of rev
V = Vo + a*T.
29.1 = 0 + a*6.5,
a = 4.48 m/s^2.
d = 0.5*a*T^2 = 0.5*4.48*6.5^2 = 94.6 m.
C = pi*2r = 3.14*0.834 = 2.62 m. = Circumference.
Rev. = 94.6/2.62 = 36.1.
a = 29.1/6.5 = 4.48 m/s^2
average speed = 29.1/2 = 14.55 m/s
d =average speed * time = 14.55 * 6.5 = 94.6 meters
circumference of tire = pi D = 3.14 * 0.834 = 2.62 meters
so
94.6 / 2.62 turns
To find the number of revolutions the tire makes during this motion, we need to find the distance traveled by the car and then convert it to the number of tire revolutions.
Step 1: Convert the diameter of the tire to the radius.
The radius, r, of the tire can be found by dividing the diameter, D, by 2.
r = D/2 = 83.4 cm / 2 = 41.7 cm
Step 2: Convert the radius of the tire to meters.
To convert the radius to meters, divide by 100 (since 1 meter = 100 cm).
r = 41.7 cm / 100 = 0.417 m
Step 3: Calculate the distance traveled by the car.
The distance traveled, s, can be calculated using the equation:
s = ut + (1/2)at^2
where u is the initial velocity (0 m/s), a is the acceleration, and t is the time.
The acceleration, a, can be found using the equation:
a = (v - u) / t
where v is the final velocity and u is the initial velocity.
Given:
Initial velocity, u = 0 m/s
Final velocity, v = 29.1 m/s
Time, t = 6.5 s
Acceleration, a = (29.1 m/s - 0 m/s) / 6.5 s = 4.47 m/s^2
Substituting the values into the distance formula:
s = (0 m/s)(6.5 s) + (1/2)(4.47 m/s^2)(6.5 s)^2
s = 0 m + (1/2)(4.47 m/s^2)(42.25 s^2)
s = (1/2)(4.47 m/s^2)(42.25 s^2)
s = 94.44 m
Step 4: Calculate the number of revolutions.
The distance traveled by one tire revolution, d, can be found using the formula:
d = 2πr
Substituting the value of the radius found earlier:
d = 2π(0.417 m)
d ≈ 2.619 m
The number of tire revolutions, N, can be found by dividing the total distance traveled by the distance per revolution:
N = s / d
N = 94.44 m / 2.619 m
N ≈ 36.05
Therefore, the number of revolutions the tire makes during this motion, assuming no slipping, is approximately 36 revolutions.
To find the number of revolutions the tire makes during this motion, we need to know the circumference of the tire.
The circumference of a circle can be calculated using the formula: circumference = π * diameter
In this case, the diameter of the tire is given as 83.4 cm. So, the circumference of the tire would be:
circumference = π * 83.4 cm
Before calculating the number of revolutions, let's convert the given speed from m/s to cm/s, since the diameter of the tire is provided in centimeters.
Given speed = 29.1 m/s
To convert this to cm/s, multiply by 100:
Converted speed = 29.1 m/s * 100 = 2910 cm/s
Now we can calculate the number of revolutions the tire makes during this motion.
Number of revolutions = distance traveled / circumference
To find the distance traveled, we can use the formula for uniformly accelerated motion:
distance = initial velocity * time + (1/2) * acceleration * time^2
In this case, the car is starting from rest, so the initial velocity is 0 m/s. The acceleration can be found using the formula:
acceleration = change in velocity / time
The change in velocity is the difference between the final speed and the initial velocity:
change in velocity = final velocity - initial velocity = 29.1 m/s
Now we can calculate the acceleration:
acceleration = 29.1 m/s / 6.5 s
Plugging in the values, we can calculate the acceleration.
acceleration = 4.47 m/s^2
Using the calculated acceleration and the given time of 6.5 s, we can calculate the distance traveled:
distance = 0 * 6.5 + (1/2) * 4.47 * (6.5)^2
Now we can calculate the number of revolutions:
Number of revolutions = distance traveled / circumference
Plugging in the values, we get:
Number of revolutions = distance / (π * 83.4 cm)
Once you calculate the distance traveled, divide it by the circumference to find the number of revolutions. Make sure to use the same units for both the distance and the circumference, such as cm or meters.